117089
If \(f\) is a relation from set of positive real numbers to the set of positive real numbers defined by \(f(x)=3 x^2-2\) then \(f\) is
1 one-one but not onto
2 onto but not one-one
3 a bijection
4 not a function
Explanation:
Exp: (D) : Given that, \(f(x)=3 x^2-2\) For one-one \(\mathrm{x}_1=\mathrm{x}_2\) \(3 \mathrm{x}_1^2-2=3 \mathrm{x}_2^2-2\) \(3 \mathrm{x}_1^2=3 \mathrm{x}_2^2\) \(\mathrm{x}_1^2=\mathrm{x}_2^2\) \(\mathrm{x}_1= \pm \mathrm{x}_2\) \(\mathrm{x}_1 \neq \mathrm{x}_2\) hence it is not one-one For onto, \(y=3 x^2-2\) \(y+2=3 x^2\) \(x^2=\frac{y+2}{3}\) \(x= \pm \sqrt{\frac{y+2}{3}}\)\(\therefore\) Hence it is not a function.
[AP EAMCET-07.07.2022
Sets, Relation and Function
117090
If a function \(\mathbf{f}: \mathbf{R}-\{\mathbf{l}\} \rightarrow \mathbf{R}-\{\mathbf{m}\}\) defined by \(f(x)=\frac{x+3}{x-2}\) is a bijection, then \(3 l+2 m=\)
1 10
2 12
3 8
4 14
Explanation:
Exp: (C) : Given, \(f(x)=\frac{x+3}{x-2}\) Domain of function is \(\mathrm{R} \rightarrow\{2\}\) Let, \(y=\frac{x+3}{x-2}\) \(x y-2 y=x+3\) \(x y-x=2 y+3\) \(x(y-1)=2 y+3\) \(x=\frac{2 y+3}{y-1}\) Range of function is \(\mathrm{R}-\{1\}\) \(\therefore \quad \quad l=2, \mathrm{~m} =1\) \(31+2 \mathrm{~m} =3 \times 2+2 \times 1\) \(=6+2\) \(=8\)
[AP EAMCET-05.07.2022
Sets, Relation and Function
117091
Function \(f: R \rightarrow R\) defined as \(f(x)=x^3\), \(f\) is
1 one-one and onto
2 one-one but not onto
3 many-one and onto
4 neither one-one nor onto
Explanation:
Exp: (A) : Given, \(f(x)=x^3\) \(\mathrm{f}: \mathrm{R} \longrightarrow \mathrm{R}\) For onto, \(\mathrm{f}\left(\mathrm{x}_1\right)=\mathrm{f}\left(\mathrm{x}_2\right) \Rightarrow \mathrm{x}_1=\mathrm{x}_2\) \(\mathrm{x}_1^3=\mathrm{x}_2^3\) \(\mathrm{x}_1^3-\mathrm{x}_2^3=0\) \(\left(\mathrm{x}_1-\mathrm{x}_2\right)\left(\mathrm{x}_1^2+\mathrm{x}_1 \mathrm{x}_2+\mathrm{x}_2^2\right)=0\) \(\mathrm{x}_1=\mathrm{x}_2\) So, \(\mathrm{f}(\mathrm{x})\) is one-one function. Also, for any real number (y) in co-domain \(\mathrm{R}\) There exists \((y)^{1 / 3}\) in \(\mathrm{R}\) such that - \(f(y)^{\frac{1}{3}}=(y)^{\frac{1}{3} \times 3}=y\) \(\therefore \mathrm{f}\) is onto. Hence, function \(f\) is one-one and onto.
[GUJCET-2021]
Sets, Relation and Function
117092
\(f: R \rightarrow R, f(x)=x^2+3 x+4\) is
1 One-one and not onto
2 Many-one and not onto
3 One-one and onto
4 not one-one and onto
Explanation:
Exp: (B) : Given, \(f(x)=x^2+3 x+4\) \(x=\frac{-3 \pm \sqrt{9-16}}{2}\) \(x=\frac{-3 \pm \sqrt{-7}}{2}\) \(\mathrm{x}=\) Imaginary Number So, function is many-one and not onto.
117089
If \(f\) is a relation from set of positive real numbers to the set of positive real numbers defined by \(f(x)=3 x^2-2\) then \(f\) is
1 one-one but not onto
2 onto but not one-one
3 a bijection
4 not a function
Explanation:
Exp: (D) : Given that, \(f(x)=3 x^2-2\) For one-one \(\mathrm{x}_1=\mathrm{x}_2\) \(3 \mathrm{x}_1^2-2=3 \mathrm{x}_2^2-2\) \(3 \mathrm{x}_1^2=3 \mathrm{x}_2^2\) \(\mathrm{x}_1^2=\mathrm{x}_2^2\) \(\mathrm{x}_1= \pm \mathrm{x}_2\) \(\mathrm{x}_1 \neq \mathrm{x}_2\) hence it is not one-one For onto, \(y=3 x^2-2\) \(y+2=3 x^2\) \(x^2=\frac{y+2}{3}\) \(x= \pm \sqrt{\frac{y+2}{3}}\)\(\therefore\) Hence it is not a function.
[AP EAMCET-07.07.2022
Sets, Relation and Function
117090
If a function \(\mathbf{f}: \mathbf{R}-\{\mathbf{l}\} \rightarrow \mathbf{R}-\{\mathbf{m}\}\) defined by \(f(x)=\frac{x+3}{x-2}\) is a bijection, then \(3 l+2 m=\)
1 10
2 12
3 8
4 14
Explanation:
Exp: (C) : Given, \(f(x)=\frac{x+3}{x-2}\) Domain of function is \(\mathrm{R} \rightarrow\{2\}\) Let, \(y=\frac{x+3}{x-2}\) \(x y-2 y=x+3\) \(x y-x=2 y+3\) \(x(y-1)=2 y+3\) \(x=\frac{2 y+3}{y-1}\) Range of function is \(\mathrm{R}-\{1\}\) \(\therefore \quad \quad l=2, \mathrm{~m} =1\) \(31+2 \mathrm{~m} =3 \times 2+2 \times 1\) \(=6+2\) \(=8\)
[AP EAMCET-05.07.2022
Sets, Relation and Function
117091
Function \(f: R \rightarrow R\) defined as \(f(x)=x^3\), \(f\) is
1 one-one and onto
2 one-one but not onto
3 many-one and onto
4 neither one-one nor onto
Explanation:
Exp: (A) : Given, \(f(x)=x^3\) \(\mathrm{f}: \mathrm{R} \longrightarrow \mathrm{R}\) For onto, \(\mathrm{f}\left(\mathrm{x}_1\right)=\mathrm{f}\left(\mathrm{x}_2\right) \Rightarrow \mathrm{x}_1=\mathrm{x}_2\) \(\mathrm{x}_1^3=\mathrm{x}_2^3\) \(\mathrm{x}_1^3-\mathrm{x}_2^3=0\) \(\left(\mathrm{x}_1-\mathrm{x}_2\right)\left(\mathrm{x}_1^2+\mathrm{x}_1 \mathrm{x}_2+\mathrm{x}_2^2\right)=0\) \(\mathrm{x}_1=\mathrm{x}_2\) So, \(\mathrm{f}(\mathrm{x})\) is one-one function. Also, for any real number (y) in co-domain \(\mathrm{R}\) There exists \((y)^{1 / 3}\) in \(\mathrm{R}\) such that - \(f(y)^{\frac{1}{3}}=(y)^{\frac{1}{3} \times 3}=y\) \(\therefore \mathrm{f}\) is onto. Hence, function \(f\) is one-one and onto.
[GUJCET-2021]
Sets, Relation and Function
117092
\(f: R \rightarrow R, f(x)=x^2+3 x+4\) is
1 One-one and not onto
2 Many-one and not onto
3 One-one and onto
4 not one-one and onto
Explanation:
Exp: (B) : Given, \(f(x)=x^2+3 x+4\) \(x=\frac{-3 \pm \sqrt{9-16}}{2}\) \(x=\frac{-3 \pm \sqrt{-7}}{2}\) \(\mathrm{x}=\) Imaginary Number So, function is many-one and not onto.
117089
If \(f\) is a relation from set of positive real numbers to the set of positive real numbers defined by \(f(x)=3 x^2-2\) then \(f\) is
1 one-one but not onto
2 onto but not one-one
3 a bijection
4 not a function
Explanation:
Exp: (D) : Given that, \(f(x)=3 x^2-2\) For one-one \(\mathrm{x}_1=\mathrm{x}_2\) \(3 \mathrm{x}_1^2-2=3 \mathrm{x}_2^2-2\) \(3 \mathrm{x}_1^2=3 \mathrm{x}_2^2\) \(\mathrm{x}_1^2=\mathrm{x}_2^2\) \(\mathrm{x}_1= \pm \mathrm{x}_2\) \(\mathrm{x}_1 \neq \mathrm{x}_2\) hence it is not one-one For onto, \(y=3 x^2-2\) \(y+2=3 x^2\) \(x^2=\frac{y+2}{3}\) \(x= \pm \sqrt{\frac{y+2}{3}}\)\(\therefore\) Hence it is not a function.
[AP EAMCET-07.07.2022
Sets, Relation and Function
117090
If a function \(\mathbf{f}: \mathbf{R}-\{\mathbf{l}\} \rightarrow \mathbf{R}-\{\mathbf{m}\}\) defined by \(f(x)=\frac{x+3}{x-2}\) is a bijection, then \(3 l+2 m=\)
1 10
2 12
3 8
4 14
Explanation:
Exp: (C) : Given, \(f(x)=\frac{x+3}{x-2}\) Domain of function is \(\mathrm{R} \rightarrow\{2\}\) Let, \(y=\frac{x+3}{x-2}\) \(x y-2 y=x+3\) \(x y-x=2 y+3\) \(x(y-1)=2 y+3\) \(x=\frac{2 y+3}{y-1}\) Range of function is \(\mathrm{R}-\{1\}\) \(\therefore \quad \quad l=2, \mathrm{~m} =1\) \(31+2 \mathrm{~m} =3 \times 2+2 \times 1\) \(=6+2\) \(=8\)
[AP EAMCET-05.07.2022
Sets, Relation and Function
117091
Function \(f: R \rightarrow R\) defined as \(f(x)=x^3\), \(f\) is
1 one-one and onto
2 one-one but not onto
3 many-one and onto
4 neither one-one nor onto
Explanation:
Exp: (A) : Given, \(f(x)=x^3\) \(\mathrm{f}: \mathrm{R} \longrightarrow \mathrm{R}\) For onto, \(\mathrm{f}\left(\mathrm{x}_1\right)=\mathrm{f}\left(\mathrm{x}_2\right) \Rightarrow \mathrm{x}_1=\mathrm{x}_2\) \(\mathrm{x}_1^3=\mathrm{x}_2^3\) \(\mathrm{x}_1^3-\mathrm{x}_2^3=0\) \(\left(\mathrm{x}_1-\mathrm{x}_2\right)\left(\mathrm{x}_1^2+\mathrm{x}_1 \mathrm{x}_2+\mathrm{x}_2^2\right)=0\) \(\mathrm{x}_1=\mathrm{x}_2\) So, \(\mathrm{f}(\mathrm{x})\) is one-one function. Also, for any real number (y) in co-domain \(\mathrm{R}\) There exists \((y)^{1 / 3}\) in \(\mathrm{R}\) such that - \(f(y)^{\frac{1}{3}}=(y)^{\frac{1}{3} \times 3}=y\) \(\therefore \mathrm{f}\) is onto. Hence, function \(f\) is one-one and onto.
[GUJCET-2021]
Sets, Relation and Function
117092
\(f: R \rightarrow R, f(x)=x^2+3 x+4\) is
1 One-one and not onto
2 Many-one and not onto
3 One-one and onto
4 not one-one and onto
Explanation:
Exp: (B) : Given, \(f(x)=x^2+3 x+4\) \(x=\frac{-3 \pm \sqrt{9-16}}{2}\) \(x=\frac{-3 \pm \sqrt{-7}}{2}\) \(\mathrm{x}=\) Imaginary Number So, function is many-one and not onto.
117089
If \(f\) is a relation from set of positive real numbers to the set of positive real numbers defined by \(f(x)=3 x^2-2\) then \(f\) is
1 one-one but not onto
2 onto but not one-one
3 a bijection
4 not a function
Explanation:
Exp: (D) : Given that, \(f(x)=3 x^2-2\) For one-one \(\mathrm{x}_1=\mathrm{x}_2\) \(3 \mathrm{x}_1^2-2=3 \mathrm{x}_2^2-2\) \(3 \mathrm{x}_1^2=3 \mathrm{x}_2^2\) \(\mathrm{x}_1^2=\mathrm{x}_2^2\) \(\mathrm{x}_1= \pm \mathrm{x}_2\) \(\mathrm{x}_1 \neq \mathrm{x}_2\) hence it is not one-one For onto, \(y=3 x^2-2\) \(y+2=3 x^2\) \(x^2=\frac{y+2}{3}\) \(x= \pm \sqrt{\frac{y+2}{3}}\)\(\therefore\) Hence it is not a function.
[AP EAMCET-07.07.2022
Sets, Relation and Function
117090
If a function \(\mathbf{f}: \mathbf{R}-\{\mathbf{l}\} \rightarrow \mathbf{R}-\{\mathbf{m}\}\) defined by \(f(x)=\frac{x+3}{x-2}\) is a bijection, then \(3 l+2 m=\)
1 10
2 12
3 8
4 14
Explanation:
Exp: (C) : Given, \(f(x)=\frac{x+3}{x-2}\) Domain of function is \(\mathrm{R} \rightarrow\{2\}\) Let, \(y=\frac{x+3}{x-2}\) \(x y-2 y=x+3\) \(x y-x=2 y+3\) \(x(y-1)=2 y+3\) \(x=\frac{2 y+3}{y-1}\) Range of function is \(\mathrm{R}-\{1\}\) \(\therefore \quad \quad l=2, \mathrm{~m} =1\) \(31+2 \mathrm{~m} =3 \times 2+2 \times 1\) \(=6+2\) \(=8\)
[AP EAMCET-05.07.2022
Sets, Relation and Function
117091
Function \(f: R \rightarrow R\) defined as \(f(x)=x^3\), \(f\) is
1 one-one and onto
2 one-one but not onto
3 many-one and onto
4 neither one-one nor onto
Explanation:
Exp: (A) : Given, \(f(x)=x^3\) \(\mathrm{f}: \mathrm{R} \longrightarrow \mathrm{R}\) For onto, \(\mathrm{f}\left(\mathrm{x}_1\right)=\mathrm{f}\left(\mathrm{x}_2\right) \Rightarrow \mathrm{x}_1=\mathrm{x}_2\) \(\mathrm{x}_1^3=\mathrm{x}_2^3\) \(\mathrm{x}_1^3-\mathrm{x}_2^3=0\) \(\left(\mathrm{x}_1-\mathrm{x}_2\right)\left(\mathrm{x}_1^2+\mathrm{x}_1 \mathrm{x}_2+\mathrm{x}_2^2\right)=0\) \(\mathrm{x}_1=\mathrm{x}_2\) So, \(\mathrm{f}(\mathrm{x})\) is one-one function. Also, for any real number (y) in co-domain \(\mathrm{R}\) There exists \((y)^{1 / 3}\) in \(\mathrm{R}\) such that - \(f(y)^{\frac{1}{3}}=(y)^{\frac{1}{3} \times 3}=y\) \(\therefore \mathrm{f}\) is onto. Hence, function \(f\) is one-one and onto.
[GUJCET-2021]
Sets, Relation and Function
117092
\(f: R \rightarrow R, f(x)=x^2+3 x+4\) is
1 One-one and not onto
2 Many-one and not onto
3 One-one and onto
4 not one-one and onto
Explanation:
Exp: (B) : Given, \(f(x)=x^2+3 x+4\) \(x=\frac{-3 \pm \sqrt{9-16}}{2}\) \(x=\frac{-3 \pm \sqrt{-7}}{2}\) \(\mathrm{x}=\) Imaginary Number So, function is many-one and not onto.
